**The Work Done to Charge a 24μF Capacitor**

When a capacitor is connected to a voltage source, energy is transferred from the source to the capacitor, causing the electric field between the plates to build up. In this article, we will calculate the work done to charge a 24μF capacitor when the potential difference between the plates is 500V.

**Capacitor Basics**

A capacitor is a device that stores electric energy in the form of an electric field. It consists of two conductive plates separated by a dielectric material, such as air, ceramic, or a polymer film. The capacitance (C) of a capacitor is the ratio of the charge (Q) stored on the plates to the potential difference (V) between them:

**C = Q / V**

**Work Done to Charge a Capacitor**

The energy (W) required to charge a capacitor is equal to the product of the charge (Q) and the potential difference (V) between the plates:

**W = QV**

Since the capacitance (C) is constant, we can rewrite the equation as:

**W = (CV)^2 / 2**

**Calculating the Work Done**

Given a 24μF capacitor and a potential difference of 500V between the plates, we can calculate the work done to charge the capacitor using the formula:

**W = (CV)^2 / 2**

Substituting the values, we get:

**W = ((24 × 10^(-6)) × 500)^2 / 2**

**W = 0.144 J**

Therefore, the work done to charge a 24μF capacitor when the potential difference between the plates is 500V is approximately **0.144 joules**.

**Conclusion**

In this article, we have calculated the work done to charge a 24μF capacitor when the potential difference between the plates is 500V. The result shows that a significant amount of energy is required to charge the capacitor, which is proportional to the capacitance, voltage, and the square of the voltage. This energy is stored in the electric field between the plates and can be released quickly when the capacitor is discharged.